Answer
$\dfrac{32}{3}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Area =\iint_D dA $
$\int_{0}^{2} \int^{0}_{x^2-4}dydx+\int_{0}^{{\sqrt{x}}} \space dy \space dx =\int_{0}^{2}(4-x^2)dx+\int_{0}^{4} \sqrt x \space dx $
or, $=[4x-\dfrac{1}{3}x^3]^2_0+[\dfrac{2x^{3/2}}{3}]^4_0$
or, $=(8-\dfrac{8}{3})+\dfrac{16}{3}$
and $ Area=\dfrac{32}{3}$
